Complete Question
The complete question is shown on the first question
Answer:
a) The duty of the heat exchanger is given as 6.8658 KJ /sec
b) The temperature of the water leaving the exchanger is TOUT = 29.84 ⁰C
c) The log mean difference is given as TZ = 47.317 ⁰ C
d) the UA value is UA = 145.10
Explanation:
The explanation is uploaded on the first and second ,third and fourth image
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
<h3>
Answer:</h3>
11.84 mol CoF₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂
[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂
[Given] 11.84 moles CoCl₂
[Solve] moles CoF₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol CoCl₂ → 1 mol CoF₂
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Answer:
The heat of vaporization is typically larger than the heat of fusion
Next question answer:
The liquid water absorbs heat from the skin surface and is transferred to the air when the water evaporates.
Explanation:
From the equation q=mCΔT, set the q of copper = to q of water,
So --- mCΔT(copper)=mCΔT(water).
mass (Cu - copper) = 38g
mass (H2O - water) = 15g
C (H2O) = 4.184 J/g*C
ΔΤ (H2O) = 33-22 = 11*C
ΔΤ (Cu) = 33-80 = -47*C (the final temp is the same for both materials - thermal equilibrium)
C (Cu) = ?
So --- 38(-47)C[Cu]=15(4.184)(11)
--- C[Cu]=690.36/(-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)
